We have that $\det T_k$ is a fixed (depending on $n=\dim V$ and $k$ only) power of
$\det T$. To see this, as well as getting the power, one can for instance note
that $\mathrm{SL}(V)$ is the commutator subgroup of $\mathrm{GL}(V)$ (except for extremely small finite fields but we can always increase the size of the field) and hence
if $\det T=1$ then $\det T_k=1$. We can then write any $T\in\mathrm{GL}(V)$ in
the form $DS$, where $S\in\mathrm{SL}(V)$ and $D$ a diagonal matrix with
diagonal entries $(t,1,1,\ldots,1)$. Then $\det((DS)_k)=\det(D_k)\cdot1$ so it
suffices to compute $\det(D_k)$ but in the standard basis of $\mathrm{Sym}^kV$,
given a basis $e_1,\ldots,e_n$ of $V$, $D_k$ is a diagonal entries and its
determinant is $t^R$, where $R=\sum_{0\leq i\leq k}is^{n-1}_{k-i}$. Here
$s^{a}_{b}=\dim \mathrm{Sym}^bU$ where $\dim U=a$ which equals $\binom{a+b-1}{b}$.
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We have that $\det T_k$ is a fixed (depending on $n=\dim V$ and $k$ only) power of
$\det T$. To see this, as well as getting the power, one can for instance note
that $\mathrm{SL}(V)$ is the commutator subgroup of $\mathrm{GL}(V)$ and hence
if $\det T=1$ then $\det T_k=1$. We can then write any $T\in\mathrm{GL}(V)$ in
the form $DS$, where $S\in\mathrm{SL}(V)$ and $D$ a diagonal matrix with
diagonal entries $(t,1,1,\ldots,1)$. Then $\det((DS)_k)=\det(D_k)\cdot1$ so it
suffices to compute $\det(D_k)$ but in the standard basis of $\mathrm{Sym}^kV$,
given a basis $e_1,\ldots,e_n$ of $V$, $D_k$ is a diagonal entries and its
determinant is $t^R$, where |
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